A non-homogeneous space such that all neighborhood spaces are homeomorphic For the sake of this question a neighborhood space is a topological space $N(x)$ which has the same point-set as a space $X$ and whose topology is the subcollection of the topology on $X$ obtained by including only those open sets containing $x$ and the empty set. 
If every neighborhood space of a space $X$ is homeomorphic, is $X$ homogeneous?
A direct proof of this might take the given neighborhood space homeomorphism taking a point $a$ to $b$ and paste to it additional neighborhood space homeomorphisms until the full space is covered, giving us a homeomorphism taking $a$ to $b$. For this to work we'd need to find neighborhood space homeomorphisms which agree on the intersection of (some of) their neighborhoods but I'm missing any simple way to do this. A proof by contradiction might proceed by supposing that $X$ is not homogeneous and so there are $a$ and $b$ such that there is no homeomorphism taking $a$ to $b$. Then the neighborhood space homeomorphism $\phi_{ab} : N(a) \to N(b)$ is a bijection of $X$ and so, without loss of generality by interchanging $a$ and $b$, there must exist some open $V$ of $X$ such that $\phi_{ab}^{-1}(V)$ is not open in $X$. If $V$ could be guaranteed to contain $b$ then we would be done but I'm missing a simple reason for why this should happen.
1-transitivity is often marketed as meaning "the thing is the same at every point", but in this case a statement which to me seems more closely aligned with this idea might not be equivalent to it. So I'm curious to see if a space exists contradicting the statement and what that space looks like.
 A: Here's a counterexample.  Let $X$ be a countably infinite set and choose a partition $$X=\bigcup_{n\in\mathbb{N}} A_n\cup \bigcup_{n\in\mathbb{N}} B_n\cup\bigcup_{n\in\mathbb{N}} C_n$$ where each $A_n,B_n,$ and $C_n$ is infinite.  Consider the topology on $X$ generated by the sets $A_n$, $A_n\cup B_n$, and $C_n$ for all $n$.  Note that $X$ is not homogeneous: for instance, if $x\in X$, then there exists a point $y\in X$ such that $x\in\overline{\{y\}}$ but $y\not\in\overline{\{x\}}$ if and only if $x\in B_n$ for some $n$ (in which case you can take $y$ to be any point of $A_n$).  However, I claim all the neighborhood spaces of $X$ are homeomorphic.
Fix points $a\in A_0$, $b\in B_0$, and $c\in C_0$.  Note that there are homeomorphisms of $X$ sending $a$ to any point in any $A_n$, or sending $b$ to any point in any $B_n$, or sending $c$ to any point in any $C_n$.  So it suffices to show that $N(a)$, $N(b)$, and $N(c)$ are homeomorphic.
First, we construct a homeomorphism $f:N(a)\to N(c)$.  We take $f$ to be any bijection $X\to X$ such that $f(A_0)=C_0$, $f(B_0)=C_1$, $f(A_n)=A_{n-1}$ for all $n>0$, $f(B_n)=B_{n-1}$ for all $n>0$, and $f(C_n)=C_{n+2}$ for all $n$.  Note that $f$ fails to be a homeomorphism $X\to X$ only because $B_0$ is not open but $f(B_0)=C_1$ is open.  However, $f$ is a homeomorphism $N(a)\to N(c)$, since to get an open set in $N(a)$ you must add in $A_0$ and $A_0\cup B_0$ is open.
Now we construct a homeomorphism $N(b)\to N(c)$.  We take $f$ to be any bijection $X\to X$ such that $f(A_0\cup B_0)=C_0$, $f(A_n)=A_{n-1}$ for all $n>0$, $f(B_n)=B_{n-1}$ for all $n>0$, and $f(C_n)=C_{n+1}$ for all $n$.  Such an $f$ fails to be a homeomorphism $X\to X$ only because $A_0$ is open but $f(A_0)$ (which is half of $C_0$) is not open.  This problem goes away when you consider $f$ as a map $N(b)\to N(c)$, since you must add $B_0$ to $A_0$ to get an open set in $N(b)$ and $f(A_0\cup B_0)$ is open.

However, there are no Hausdorff counterexamples.  In fact, something stronger is true:

Let $X$ be a Hausdorff space, $a,b\in X$, and $f:N(a)\to N(b)$ be a homeomorphism.  Then $f(a)=b$ and $f$ is also a homeomorphism as a map $X\to X$.

First, $a$ is the unique point of $N(a)$ that is contained in all nonempty open sets and similarly for $b$ in $N(b)$, so $f(a)=b$.  To prove $f:X\to X$ is a homeomorphism, it suffices to show $f$ is continuous, since the same reasoning will show $f^{-1}$ is continuous as well.
Suppose $U\subseteq X$ is open.  If $b\in U$, then $U$ is open in $N(b)$, so $f^{-1}(U)$ is open in $N(a)$ and hence in $X$.  So we may assume $b\not\in U$.  Let $x\in f^{-1}(U)$.  Since $b\not\in U$, $x\neq a$.
Since $X$ is Hausdorff, there exists an open neighborhood $V$ of $a$ in $X$ whose closure does not contain $x$.  Since $V$ is open in $N(a)$, $f(V)$ is open in $N(b)$ and hence in $X$. Now observe that $U\cup f(V)$ is open in $N(b)$, and therefore $f^{-1}(U\cup f(V))=f^{-1}(U)\cup V$ is open in $N(a)$ and hence in $X$.  Therefore $(f^{-1}(U)\cup V)\setminus \overline{V}=f^{-1}(U)\setminus\overline{V}$ is also open in $X$ (here the closure is computed in $X$).  By our choice of $V$, $x\in f^{-1}(U)\setminus\overline{V}$.  Thus $f^{-1}(U)\setminus\overline{V}$ is an open neighborhood of $x$ in $X$ which is contained in $f^{-1}(U)$.
Since $x\in f^{-1}(U)$ was arbitrary, this proves that $f^{-1}(U)$ is open in $X$.  Thus $f$ is continuous from $X$ to $X$, as desired.
